How do you factor #y^2-16+64#?

1 Answer
Jan 24, 2016

Answer:

This trinomial is a perfect square trinomial, so can be factored as #(a-b)^2#

Explanation:

The square root of 64 is 8 and the square root of #y^2# is y.

The middle term should be of the form 2ab. In this case, a=y and b= 8

So, 2(y)(8), or 16y, which is the middle term. This short proof justifies that it is indeed a perfect square trinomial.

#y^2# - 16y + 64
= (y - 8)(y - 8)
= #(y - 8)^2#

Your answer is #(y -8)^2#

Exercises:

  1. Out of the following trinomials, which is/are perfect square trinomial(s)?

a). #y^2# - 16

B) #y^2# + 8y + 16

C) #y^2# + 16

D) #y^2# - 8y + 16

  1. Factor the following perfect square trinomial:

#4x^2# + 28x + 49

Hopefully you understand now!